Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 30)
30.
What will be remainder when (6767 + 67) is divided by 68 ?
Answer: Option
Explanation:
(xn + 1) will be divisible by (x + 1) only when n is odd.
(6767 + 1) will be divisible by (67 + 1)
(6767 + 1) + 66, when divided by 68 will give 66 as remainder.
Discussion:
70 comments Page 1 of 7.
Ummar said:
3 years ago
@All.
Here is the solution;
67 * 2 - 68 = remainder = 66.
Valid for any type of formation like the given one.
n*2-(n+1) = remainder.
Here is the solution;
67 * 2 - 68 = remainder = 66.
Valid for any type of formation like the given one.
n*2-(n+1) = remainder.
(34)
Gnanam said:
2 years ago
1 IS NOT DIVISIBLE BY 6 & 8.
(6+3) =9 -> 9 is not divisible by 6 & 8.
(6+7) -> 13 is not divisible by 6 & 8.
(6+6) -> 12 is divisible by 6 & 8 ===> ANSWER.
(6+3) =9 -> 9 is not divisible by 6 & 8.
(6+7) -> 13 is not divisible by 6 & 8.
(6+6) -> 12 is divisible by 6 & 8 ===> ANSWER.
(17)
Sumit said:
9 months ago
(67^67+67)/68,
n(1^n +1)/68,
67(1^67+1)/68,
67(1+1)/68 = give reminder 66.
n(1^n +1)/68,
67(1^67+1)/68,
67(1+1)/68 = give reminder 66.
(9)
Kamal hachan said:
5 years ago
(67^67+67)%68 where % is mod operator. it gives remainder.
= 67^67%68+67%68
=(68-1)^67%68 +(68-1)%68
=(-1)^67+(-1)
= -1-1
=-2 (Since remainder can't be negative )
=68-2
= 66
believe me this is best trick for remainder
If any doubt please email me.
= 67^67%68+67%68
=(68-1)^67%68 +(68-1)%68
=(-1)^67+(-1)
= -1-1
=-2 (Since remainder can't be negative )
=68-2
= 66
believe me this is best trick for remainder
If any doubt please email me.
(8)
Saikumar Padi said:
4 years ago
Here, we can use n^n+n = (n+1) (n^n+1)/(n+1) + (n-1).
As division n=p*q+r.
As division n=p*q+r.
(7)
Bhaskar said:
5 years ago
By using binomial theorem.
((68-1)^67+67)/68.
All term which includes 68 will be divided by 68.
Then((-1)^67+67)/68 will be left,
It becomes 66/68.
Hence remainder is 66.
((68-1)^67+67)/68.
All term which includes 68 will be divided by 68.
Then((-1)^67+67)/68 will be left,
It becomes 66/68.
Hence remainder is 66.
(6)
Pushpendra Shukla said:
4 years ago
(x^n - a^n) is divisible by (x - a) for all values of n.
(x^n - a*n) is divisible by (x + a) for even values of n.
(x^n + a^n) is divisible by (x + a) for odd values of n.
(x^n - a*n) is divisible by (x + a) for even values of n.
(x^n + a^n) is divisible by (x + a) for odd values of n.
(5)
Suma said:
5 years ago
I didn't understand, Anyone Help me out.
(4)
Barath said:
4 years ago
Using unit digit method : 67 power 67 is 343 and we will add 343 +67.
410 and it is divided by 68 then the remainder will be 2.
410 and it is divided by 68 then the remainder will be 2.
(4)
Mahi said:
1 decade ago
Simply if you got dis situation do as follow
jus 67*2=134
if u hav any doubt check wid another simple example as follow
3pwr3+3 check wedr it is divisible with 4 r not
by solving in normal way we will get as follow
3pwr3+3 you will get 30. divide 30 with 4 you will get 2 reminder
according to our procedure
3*2=6
So better to go this way guys.
jus 67*2=134
68)134(1
68
---
66
if u hav any doubt check wid another simple example as follow
3pwr3+3 check wedr it is divisible with 4 r not
by solving in normal way we will get as follow
3pwr3+3 you will get 30. divide 30 with 4 you will get 2 reminder
according to our procedure
3*2=6
4)6(1
4
--
2
So better to go this way guys.
(3)
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